First, some notation. Let us represent an element of $\mathbb{Z}/p^n\mathbb{Z}$ by the pair $(a,n)$, where $a\in\mathbb{Z}$; so $(a,n)=(b,n)$ if and only if $a\equiv b\pmod{p^n}$.
We write $\mathbb{Z}_{p^{\infty}}$ for the direct limit; the elements of the direct limit are equivalence classes of the form $[a,n]$, $a\in\mathbb{Z}$, $n\geq 1$, with $[a,n]=[b,m]$ if and only if (i) $(p^{m-n}a,m)=(b,m)$ if $m\geq n$; or (ii) $(a,n) = (p^{n-m}b,n)$ if $m\lt n$. Addition is defined by
$$[a,n]+[b,m] = \left\{\begin{array}{ll}
\ [a+p^{n-m}b,n] &\text{if }n\geq m;\\
\ [p^{m-n}a+b,m] &\text{if }n\lt m.
\end{array}\right.$$
The universal property of the direct limit means that if $G$ is any group, and we have group homomorphisms $f_n\colon \mathbb{Z}/p^n\mathbb{Z}\to G$ such that for all $n\leq m$, $f_n(a,n) = f_m(p^{m-n}a,m)$, then there exists a unique $f\colon\mathbb{Z}_{p^{\infty}}\to G$ such that $f[a,n] = f_n(a)$.
First, we need to show that $\mathbb{Z}_{p^{\infty}}$ satisfies the conditions given; then we want to show that given any group $G$ that satisfies the given conditions, we have an isomorphism from $G$ to $\mathbb{Z}_{p^{\infty}}$ (alternatively, show that $G$ has the universal property associated to the direct limit).
Property 1. The unique infinite $p$-group in which every element has exactly $p$ distinct $p$th roots.
Note that $\mathbb{Z}_{p^{\infty}}$ is infinite (it contains at least $p^n$ elements for each $n$. Also, each element has order a power of $p$, since $p^n[a,n] = [p^na,n] = [0,n]$ (since $p^na\equiv 0\pmod{p^n}$). So $\mathbb{Z}_{p^{\infty}}$ is an infinite $p$-group.
Also, every element of $\mathbb{Z}_{p^{\infty}}$ has at least $p$ distinct $p$th roots: let $[a,n]\in\mathbb{Z}_{p^{\infty}}$. Then for each $k$, $0\leq k\lt p$, we have
$$p[a+kp^n,n+1] = [pa+kp^{n+1},n+1] = [pa,n+1] = [a,n];$$
and $[a+kp^n,n+1] = [a+\ell p^n,n+1]$ if and only if $kp^n\equiv \ell p^n\pmod{p^{n+1}}$, if and only if $k\equiv \ell\pmod{p^n}$. Since we are assuming $0\leq k,\ell\lt p$, we conclude that $k=\ell$, so $[a,n]$ has at least $p$ distinct $p$th roots.
To show there are exactly $p$ distinct $p$th roots, first we show there are exactly $p$ distinct elements of order dividing $p$. If $p[b,m] = [0,m]$, then $pb\equiv 0\pmod{p^m}$, hence $p^{m-1}|b$. Writing $b=p^{m-1}k$ we have $[b,m] = [kp^{m-1},m] = [k,1]$. Thus, every element of order $p$ must be one of $[k,1]$ with $0\leq k\lt p$, so there are at most $p$ elements of order dividing $p$; each of these is distinct, so there are exactly $p$ elements of order dividing $p$.
Now suppose that $[b,m]$ is a $p$th root of $[a,m]$. Then for each $k$, $0\leq k\lt p$, we have
$$p\Bigl( [b,m] - [a+kp^n,n+1]\Bigr) = 0.$$
Therefore,
$$\Bigl\{ [b,m] - [a+kp^n,n+1]\Bigm| k=0,1,\ldots,p-1\Bigr\} = \Bigl\{ [0,1], [1,1],\ldots,[p-1,1]\Bigr\}$$
(since the $[b,m]-[a+kp^n,n+1]$ are pairwise distinct and all elements of order $p$). Therefore, there exists $k$ such that $[b,m]-[a+kp^n,n+1] = [0,1]$, hence $[b,m]=[a+kp^n,n+1]$, proving that the aforementioned $p$th roots of $[a,n]$ are the only $p$th roots of $[a,n]$. Thus, $[a,n]$ has exactly $p$ distinct $p$th roots. This proves that $\mathbb{Z}_{p^{\infty}}$ is an infinite $p$-group in which every element has exactly $p$ distinct $p$th roots.
The next step is to show that any infinite $p$-group in which every element has exactly $p$ distinct $p$th roots is isomorphic to the direct limit. Let $G$ be such a group.
First, we construct maps $f_m\colon \mathbb{Z}/p^m\mathbb{Z}\to G$ for each $m$ inductively, ensuring at each step that the maps form a consistent system. Since we do not know ahead of time whether $G$ is abelian, I will use multiplicative notation for $G$. For each $n\in\mathbb{N}$, let
$$G[p^n] = \{g\in G\mid g^{p^n}=1\}.$$
Note that $G[p]\subseteq G[p^2]\subseteq G[p^3]\subseteq\cdots$.
Consider $G[p] = \{g\in G\mid g^p = 1\}$. We know that $G[p]$ contains $p$ distinct elements (since by assumption, $1$ has exactly $p$ distinct $p$th roots in $G$). Let $x_1\in G[p]$, $x_1\neq 1$. Then $\langle x_1\rangle$ is cyclic of order $p$, and every element is of exponent $p$. Therefore, $G[p]=\langle x_1\rangle$. Define $f_1\colon \mathbb{Z}/p\mathbb{Z}\to \langle x_1\rangle$ by $f_1(1,1) = x_1$.
Assume that we have shown that $G[p^k]$ is a cyclic group of order $p^k$, generated by $x_k$ for $1\leq k\leq n$, and have defined maps $f_k\colon\mathbb{Z}/p^k\mathbb{Z} \to G[p^k]$ in such a way that the maps commute with the natural embeddings $\mathbb{Z}/p^k\mathbb{Z}\hookrightarrow \mathbb{Z}/p^{k+1}\mathbb{Z}$ for all $k$, $0\leq k\lt n$. In particular, we have that $x_{k+1}^p = x_k$ for each $k$.
Consider $G[p^{n+1}]$. Note that $g\in G[p^{n+1}]$ if and only if $g^p\in G[p^n]$. Since $G[p^n]$ has $p^n$ elements, and each element has exactly $p$ distinct $p$th roots, we conclude that $G[p^{n+1}]$ has $p\times p^n = p^{n+1}$ elements. Let $x_{n+1}$ be a $p$-th root of $x_n$. Then $x_{n+1}$ has order $p^{n+1}$ (since $x_n$ has order $p^n$), and since $\langle x_{n+1}\rangle \subseteq G[p^{n+1}]$, and both have $p^{n+1}$ elements, then $G[p^{n+1}]$ is a cyclic group, generated by $x_{n+1}$. Define $f_{n+1}\colon\mathbb{Z}/p^{n+1}\mathbb{Z}\to G[p^{n+1}]$ by mapping $[1,n+1]$ to $x_{n+1}$. This map is compatible with the embedding $\mathbb{Z}/p^n\mathbb{Z}\hookrightarrow \mathbb{Z}/p^n\mathbb{Z}$, and hence with all the previous embeddings as well. This completes the induction step.
Note that since $G$ is a $p$-group, if $g\in G$ then there exists $k$ such that $g\in G[p^k]$; thus, $G = \cup G[p^k]$.
Thus, we have a family of consistent maps $f_m\colon\mathbb{Z}/p^m\mathbb{Z}\to G$.
To show that $G$ has the universal property of the direct limit, let $H$ be any group, and let $h_m\colon\mathbb{Z}/p^m\mathbb{Z}\to H$ be a family of group homomorphisms which commute with the embeddings $\mathbb{Z}/p^k\mathbb{Z}\hookrightarrow \mathbb{Z}/p^{k+1}\mathbb{Z}$.
We define $h\colon G\to H$. Let $g\in G$; let $k$ be an integer such that $g\in G[p^k] = \langle x_k\rangle$. Write $g = x_k^r$, and define $h(g) = h_k(r,k)$.
Note that $h$ is well-defined: if $g = x_k^r = x_m^s$, with $k\leq m$, then we know that $x_k = x_m^{p^{m-k}}$, hence $x_k^r = x_m^{rp^{m-k}}$. Hence $s\equiv rp^{m-k}\pmod{p^m}$ (since $x_m$ is of order $p^m$). Therefore,
$$h_k(r,k) = h_m(p^{m-k}r,m) = h_m(s,m)$$
which shows $h(g)$ is well defined (we have used that the $h_n$ are a consistent system, so that $h_k(r,k) = h_{k+1}(pr,k+1)$, etc).
Also, $h$ is a group homomorphism: if $g,g'\in G$, let $k$ be such that $g,g'\in G[p^k]$. Then $gg'\in G[p^k]$ (since $G[p^k]$ is a subgroup), and so
$$h(gg') = h_k(gg') = h_k(g)h_k(g') = h(g)h(g').$$
And for every $m$, we have $h_m(a,m) = h(f_m(a,m))$ by definition, so $h$ fits into the required commutative diagrams.
Now assume that $\psi\colon G\to H$ is any group homomorphism such that $h_m = \psi\circ f_m$ for each $m$. Then $\psi(x_k) = \psi(f_k(1,k)) = h_k(1,k) = h(x_k)$. Since $G$ is generated by the $x_k$, this proves that $\psi=h$.
Thus, $G$ has the properties of the direct limit of the system
$$\mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}/p^2\mathbb{Z}\to\mathbb{Z}/p^3\mathbb{Z}\to\cdots$$
and therefore there exists a (unique) isomorphism $\mathbb{Z}_{p^{\infty}}\cong G$. Thus, $\mathbb{Z}_{p^{\infty}}$ is the unique (up to unique isomorphism) infinite $p$-group in which every element has exactly $p$ distinct $p$th roots, as claimed. QED
Property 2. The unique infinite $p$ group that is locally cyclic.
We already know that $\mathbb{Z}_{p^{\infty}}$ is an infinite $p$-group. To show it is locally cyclic, it suffices to show that any two elements are contained in a cyclic subgroup (since then the subgroup they generate is itself cyclic).
Let $[a,m]$ and $[b,n]$ be elements of $\mathbb{Z}_{p^{\infty}}$, with $m\leq n$. Then $[a,m] = [p^{n-m}a,n] = p^{n-m}a[1,n]$, and $[b,n]=b[1,n]$. Hence,
$$[a,m],[b,n] \in\langle [1,n]\rangle.$$
Thus, $\mathbb{Z}_{p^{\infty}}$ is locally cyclic.
Now let $G$ be a locally cyclic infinite $p$-group. We will show that every element of $G$ has exactly $p$ distinct $p$th roots, which will establish the desired isomorphism using part 1.
Let $a\in G$, and let $|\langle a\rangle|=p^n$. Since $a$ is of finite order and $G$ is infinite, there exists $b\in G$, $b\notin \langle a\rangle$. In particular, the subgroup generated by $a$ and $b$ is cyclic, $\langle a,b\rangle = \langle c\rangle$, and $|\langle c\rangle| \gt p^n$. Thus, $\langle c\rangle\cong \mathbb{Z}/p^{n+k}\mathbb{Z}$ for some $k\gt 0$; in this subgoup $a$ has exactly $p$ distinct $p$th roots. Thus, every element of $G$ has at least $p$ distinct $p$th roots.
Now suppose that $r_1,\ldots,r_p$ are $p$ distinct $p$th roots of $a$, and let $s$ be any $p$th root of $a$. Then $\langle r_1,\ldots,r_p,s\rangle$ is cyclic of order a power of $p$, and contains $a$. Since in a cyclic group of order $p^n$ each element has at most $p$ distinct $p$th roots, then $s$ must be one of $r_1,\ldots,r_p$, proving that $a$ has exactly $p$ distinct $p$th roots.
Thus, each element of $G$ has exacty $p$ distinct $p$th roots, and so is isomorphic to $\mathbb{Z}_{p^{\infty}}$ by the first part. QED
Best Answer
Yes, send $a/b \in \mathbb{Q}/\mathbb{Z}$ to $\zeta_{b}^{a}$, where $\zeta_{b}$ is a primitive $b$-th root of unity.